Triangles and Congruency
The perimeter of ΔEPU = 2 × (c+ r )
Step-by-step explanation:
Given OM = r and UP = c and OM⊥EP
⇒ EM = r ( As OM ⊥EP, OM = r So OM = EM)
If we draw a line joining OP , then we get two triangles ΔOMP and ΔOBC
where B is point on UP and OB is radius
Both ΔOMP and ΔOBC will be congruent, OM and OB are radius.
EP = EM + MP
⇒ MP = EP -r
If we draw a line joining OU , then we get two triangles ΔOUB and ΔOUA
where A is point on UE and OA is radius
Both ΔOUA and ΔOUB will be congruent, OA and OB are radius.
UE = UA + AE
EA = r ( As OA ⊥ UE, OM = r So OA= EM)
UP = UB + BP
c = UB + BP
Since ΔOUA and ΔOUB are congruent and ΔOMP and ΔOBC are congruent
⇒ UB = UA and MP = BP
⇒ c = UA + MP
so the perimeter of ΔEPU = EU + UP + EP
⇒ Perimeter = EU + c + EP
⇒ Perimeter = UA + AE + c + EM + MP
⇒ Perimeter = UA + r + c + r + MP
⇒ Perimeter = c+ r + c + r
⇒ Perimeter = 2 × (c+ r )
Hence , the perimeter of ΔEPU = 2 × (c+ r )
